How can the $\lambda$-return be defined recursively?
To rewrite $G_t^\lambda$ recursively, our goal is to define it in terms of
$$G_{t+1}^\lambda = (1-\lambda)\sum_{n=1}^\infty \lambda^{n-1}G_{t+1:t+n+1}.\tag{0}$$

The $\lambda$-return is a weighted average of all $n$-step returns. We will split up the summation by  pulling out the one-step return $G_{t:t+1}$ and the first step's reward $R_{t+1}$.

$$
\begin{align*}
G_t^\lambda &amp;= (1-\lambda)\sum_{n=1}^\infty \lambda^{n-1}G_{t:t+n} \tag{1}\\
&amp;\\
&amp;= (1-\lambda)\lambda^0G_{t:t+1} + (1-\lambda)\sum_{n=2}^\infty \lambda^{n-1}G_{t:t+n}\tag{2}\\
&amp;\\
&amp;= (1-\lambda)\left(R_{t+1}+\gamma\hat{v}(S_{t+1})\right)\\
&amp;\qquad + (1-\lambda)\sum_{n=2}^\infty \lambda^{n-1}(R_{t+1}+\gamma R_{t+2}+\dots +\gamma^{n-1}R_{t+n} + \gamma^n\hat{v}(S_{t+n}))\tag{3}\\
&amp;\\
&amp;= (1-\lambda)\left(R_{t+1}+\gamma\hat{v}(S_{t+1})\right) + (1-\lambda)\sum_{n=2}^\infty \lambda^{n-1} R_{t+1}\\
&amp;\qquad + (1-\lambda)\sum_{n=2}^\infty \lambda^{n-1}(\gamma R_{t+2}+\dots +\gamma^{n-1}R_{t+n} + \gamma^n\hat{v}(S_{t+n}))\tag{4}\\
&amp;\\
&amp;= \gamma(1-\lambda)\hat{v}(S_{t+1}) + (1-\lambda)\sum_{n=1}^\infty \lambda^{n-1} R_{t+1}\\
&amp;\qquad + (1-\lambda)\sum_{n=2}^\infty \lambda^{n-1}(\gamma R_{t+2}+\dots +\gamma^{n-1}R_{t+n} + \gamma^n\hat{v}(S_{t+n}))\tag{5}\\
&amp;\\
&amp;= \gamma(1-\lambda)\hat{v}(S_{t+1}) + R_{t+1}\\
&amp;\qquad + (1-\lambda)\sum_{n=2}^\infty \lambda^{n-1}(\gamma R_{t+2}+\dots +\gamma^{n-1}R_{t+n} + \gamma^n\hat{v}(S_{t+n}))\tag{6}\\
&amp;\\
&amp;= \gamma(1-\lambda)\hat{v}(S_{t+1}) + R_{t+1}\\
&amp;\qquad + \gamma\lambda(1-\lambda)\sum_{n=2}^\infty \lambda^{n-2}(R_{t+2}+\dots +\gamma^{n-2}R_{t+n} + \gamma^{n-1}\hat{v}(S_{t+n}))\tag{7}\\
&amp;\\
&amp;= \gamma(1-\lambda)\hat{v}(S_{t+1}) + R_{t+1}\\
&amp;\qquad + \gamma\lambda(1-\lambda)\sum_{m=1}^\infty \lambda^{m-1}(R_{t+2}+\dots +\gamma^{m-1}R_{t+m+1} + \gamma^{m}\hat{v}(S_{t+m+1}))\tag{8}\\
&amp;\\
&amp;= \gamma(1-\lambda)\hat{v}(S_{t+1}) + R_{t+1} + \gamma\lambda(1-\lambda)\sum_{m=1}^\infty \lambda^{m-1}G_{t+1:t+m+1}\tag{9}\\
&amp;\\
&amp;= \gamma(1-\lambda)\hat{v}(S_{t+1}) + R_{t+1} + \gamma\lambda G_{t+1}^\lambda \tag{10}\\
\end{align*}
$$

$ $
$(2)$ pulls out the one-step return from the summation.
$(3)$ expands the $n$-step returns.
$(4)$ pulls out the remaining first step rewards.
$(5)$ combines first step rewards.
$(6)$ simplifies the geometric series.
$(7)$ pulls a factor of $\gamma\lambda$ out of the summation.
$(8)$ makes the substitution $m=n-1$.
$(9)$ uses the definition of the $n$-step return.
$(10)$ uses the definition of the $\lambda$-return

The result can be verified in equation $(12.18)$ of Sutton and Barto's RL book.